THE HAHN-BANACH THEOREM - NUMERICAL PROBLEMS
AND APPLICATIONS
1 NUMERICAL PROBLEMS ON THE HAHN-BANACH THEOREM
1. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
2. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
3. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
4. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
5. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
6. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
7. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
8. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
9. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
10. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
11. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
12. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
13. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
14. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
15. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
16. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
17. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
18. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
19. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
20. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
21. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
22. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
23. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
24. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
25. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
26. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
27. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
28. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
29. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
30. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
31. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
32. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
33. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
34. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
35. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
36. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
37. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
38. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
39. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
40. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
41. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
42. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
43. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
44. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
45. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
46. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
47. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
48. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
49. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
50. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
51. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
52. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
53. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
54. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
55. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
56. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
57. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
58. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
59. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
60. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
61. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
62. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
63. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
64. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
65. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
66. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
67. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
68. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
69. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
70. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
71. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
72. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
73. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
74. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
75. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
76. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
77. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
78. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
79. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
80. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
81. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
82. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
83. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
84. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
85. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
86. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
87. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
88. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
89. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
90. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
91. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
92. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
93. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
94. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
95. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
96. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
97. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
98. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
99. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
100. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
101. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
102. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
103. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
104. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
105. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
106. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
107. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
108. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
109. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
110. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
111. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
112. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
113. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
114. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
115. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
116. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
117. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
118. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
119. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
120. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
121. Consider the vector space 𝐶[0,1] of continuous functions on [0,1] with the supremum
norm. Let 𝑌 be the subspace of polynomials of degree at most 2, and define 𝑓:𝑌→ℝ by
𝑓(𝑝)=𝑝(1/2). Find an extension 𝐹:𝐶[0,1]→ℝ of 𝑓 with ∥𝐹∥=∥𝑓∥.
Solution: First, we need to find ∥𝑓∥. For any 𝑝∈𝑌,
|𝑓(𝑝)|=|𝑝(1/2)|≤ sup
𝑥∈[0,1]|𝑝(𝑥)|=∥𝑝∥
So ∥𝑓∥≤1. The equality is achieved for 𝑝(𝑥)=1, so ∥𝑓∥=1.
By the Hahn-Banach theorem, there exists an extension 𝐹:𝐶[0,1]→ℝ with ∥𝐹∥=∥𝑓∥=
1.
One such extension is given by the point evaluation functional:
𝐹(𝑔)=𝑔(1/2) for all 𝑔∈𝐶[0,1]
122. Let 𝑋=ℓ1 and 𝑌={(𝑥𝑛)∈ℓ1:𝑥𝑛=0 for 𝑛 odd}. Define 𝑓:𝑌→ℝ by 𝑓((𝑥𝑛))=∑𝑛 even
𝑥𝑛
𝑛. Find ∥𝑓∥ and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥𝑛)∈𝑌,
|𝑓((𝑥𝑛))|=|∑𝑥𝑛
𝑛
𝑛 even |≤ ∑ |𝑥𝑛|
𝑛
𝑛 even ≤1
2∑|𝑥𝑛|
𝑛 even =1
2∥(𝑥𝑛)∥
So ∥𝑓∥≤1
2. This bound is achieved for (𝑥𝑛) with 𝑥2=1 and 𝑥𝑛=0 for 𝑛≠2.
An extension 𝐹:ℓ1→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹((𝑥𝑛))= ∑ 𝑥𝑛
𝑛
𝑛 even
123. In ℝ3 with the Euclidean norm, let 𝑌={(𝑥,𝑦,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by
𝑓(𝑥,𝑦,0)=2𝑥+𝑦. Extend 𝑓 to 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0)∈𝑌,
|𝑓(𝑥,𝑦,0)|=|2𝑥+𝑦|≤√4𝑥2+𝑦2≤√4(𝑥2+𝑦2)=2√𝑥2+𝑦2=2∥(𝑥,𝑦,0)∥
So ∥𝑓∥≤2. This bound is achieved for (1,0,0), so ∥𝑓∥=2.
An extension 𝐹:ℝ3→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑥,𝑦,𝑧)=2𝑥+𝑦
124. Let 𝑋=𝐶[0,1] with the 𝐿1 norm ∥𝑓∥1=∫|𝑓(𝑡)|
1
0𝑑𝑡. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=𝑎+2𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|𝑎+2𝑏|≤|𝑎|+2|𝑏|≤∫|𝑎𝑡+𝑏|
1
0𝑑𝑡+∫ |𝑏|
1
0𝑑𝑡=∥𝑝∥1
So ∥𝑓∥≤1. This bound is achieved for 𝑝(𝑡)=1, so ∥𝑓∥=1.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1 is given by:
𝐹(𝑔)=∫𝑔
1
0(𝑡)𝑑𝑡+𝑔(0)
125. In ℓ∞, let 𝑌={(𝑥𝑛):𝑥𝑛=0 for 𝑛>2} and define 𝑓:𝑌→ℝ by 𝑓((𝑥1,𝑥2,0,…))=3𝑥1−
2𝑥2. Find an extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥1,𝑥2,0,…)∈𝑌,
|𝑓((𝑥1,𝑥2,0,…))|=|3𝑥1−2𝑥2|≤3|𝑥1|+2|𝑥2|≤3∥(𝑥1,𝑥2,0,…)∥∞
So ∥𝑓∥≤3. This bound is achieved for (1,0,0,…), so ∥𝑓∥=3.
An extension 𝐹:ℓ∞→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹((𝑥𝑛))=3𝑥1−2𝑥2
126. Let 𝑋=𝐶[0,1] with the supremum norm, and 𝑌={𝑓∈𝑋:𝑓(0)=𝑓(1)=0}. Define
𝑓:𝑌→ℝ by 𝑓(𝑔)=∫𝑡
1
0𝑔(𝑡)𝑑𝑡. Extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑔∈𝑌,
|𝑓(𝑔)|=|∫𝑡
1
0𝑔(𝑡)𝑑𝑡|≤∫ 𝑡
1
0|𝑔(𝑡)|𝑑𝑡≤∥𝑔∥∞∫ 𝑡
1
0𝑑𝑡=1
2∥𝑔∥∞
So ∥𝑓∥≤1
2. This bound is achieved for 𝑔(𝑡)=sin(𝜋𝑡), so ∥𝑓∥=1
2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=1
2 is given by:
𝐹(ℎ)=∫ 𝑡
1
0ℎ(𝑡)𝑑𝑡−1
2ℎ(0)
127. In ℝ4 with the ℓ1 norm, let 𝑌={(𝑥,𝑦,0,0):𝑥,𝑦∈ℝ} and define 𝑓:𝑌→ℝ by 𝑓(𝑥,𝑦,0,0)=
2𝑥−3𝑦. Extend 𝑓 to 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For (𝑥,𝑦,0,0)∈𝑌,
|𝑓(𝑥,𝑦,0,0)|=|2𝑥−3𝑦|≤2|𝑥|+3|𝑦|=∥(𝑥,𝑦,0,0)∥1
So ∥𝑓∥≤3. This bound is achieved for (0,−1,0,0), so ∥𝑓∥=3.
An extension 𝐹:ℝ4→ℝ with ∥𝐹∥=∥𝑓∥=3 is given by:
𝐹(𝑥,𝑦,𝑧,𝑤)=2𝑥−3𝑦
128. Let 𝑋=𝐶[0,1] with the 𝐿2 norm ∥𝑓∥2=(∫|𝑓(𝑡)|2
1
0𝑑𝑡)1/2. Let 𝑌 be the subspace of
polynomials of degree at most 1, and define 𝑓:𝑌→ℝ by 𝑓(𝑎𝑡+𝑏)=2𝑎−𝑏. Find ∥𝑓∥
and extend 𝑓 to 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥.
Solution: For 𝑝(𝑡)=𝑎𝑡+𝑏∈𝑌,
|𝑓(𝑝)|=|2𝑎−𝑏|≤√4𝑎2+𝑏2≤√4∫ (𝑎𝑡+𝑏)2
1
0𝑑𝑡=2∥𝑝∥2
So ∥𝑓∥≤2. This bound is achieved for 𝑝(𝑡)=2𝑡−1, so ∥𝑓∥=2.
An extension 𝐹:𝑋→ℝ with ∥𝐹∥=∥𝑓∥=2 is given by:
𝐹(𝑔)=2∫ 𝑔
1
0(𝑡)𝑑𝑡−𝑔(0)
